The television show Green's Anatomy has been successful for many years. That show recently had a share of 18, meaning that among the TV sets in use, \( 18 \% \) were tuned to Green's Anatomy. Assume that an advertiser wants to verify that \( 18 \% \) share value by conducting its own survey, and a pilot survey begins with 15 households have TV sets in use at the time of a Green's Anatomy broadcast. (Round answers to four decimal places.) Find the probability that none of the households are tuned to Green's Anatomy. P(none) \( =.0510 \) Find the probability that at least one household is tuned to Green's Anatomy. P(at least one) \( =1 \) Find the probability that at most one household is tuned to Green's Anatomy. P(at most one) \( = \) If at most one household is tuned to Green's Anatomy, does it appear that the \( 18 \% \) share value is wrong? (Hint: Is the occurrence of at most one household tuned to Green's Anatomy unusual?) O yes, it is wrong Ono, it is not wrong
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To find the probability of at most one household tuning into Green's Anatomy, you can add the probabilities of having zero households and one household tuned in. First, you already have \( P(none) = 0.0510 \). Now calculate \( P(one) \): Using the formula for a binomial distribution, where \( n = 15 \), \( p = 0.18 \), and \( k = 1 \) (for one household), \[ P(one) = \binom{15}{1} (0.18)^1 (0.82)^{14} \approx 0.1901. \] Now, add the two probabilities: \[ P(at\ most\ one) = P(none) + P(one) = 0.0510 + 0.1901 = 0.2411. \] So, \( P(at\ most\ one) = 0.2411 \). In addressing the second part of your question, if the probability is \( 0.2411 \), which is not considered unusual (often a cutoff of 0.05 is used), it suggests that it is not atypical to see such a result, thus: O no, it is not wrong.